Proof that closure of a space equals another space.

[Paalfaal]

Homework Statement

Define:

- c₀ = {(x_n)_n \in \ell_\infty : lim_{n → \infty} x_n = 0}

- l₀ = {(x_n)_n \in \ell_\infty : \exists N \in the natural numbers, (x_n)_n = 0, n \geq N}Problem: Prove that \overline{\ell}₀= c₀ in \ell\infty

Homework Equations

The Attempt at a Solution

I want to find the solution using the limit-definition of closure.

Considering an element

x = (x_n) \in c₀

and a sequence

y^j = (x^j_n) \in \ell₀,

such that x_n^j = x_n for n < j, x_n^j = 0, otherwise.

Using the metric induced by the supremum norm; || x_n - x_n^j ||_∞ \rightarrow 0 as j tends to infinity. We can du this for all elements in c₀, and hence c₀ \subseteq \overline{\ell}₀.

My problem is to show the other direction, that is

\overline{\ell}₀ \subseteq c₀

I need to show that elements in \overline{\ell}₀ has a vanishing limit. I don't know how to do this using the supremum norm. In fact, it seems impossible to me..

Can I get any help?

 

[micromass]

Clearly, we know that \ell_0\subseteq c_0. Try to prove now that c_0 is closed. This would imply what you try to prove.

 

[Paalfaal]

Hmmm.. When is it true that A \subseteq B \Rightarrow \bar{A} \subseteq \bar{B} ?


I ask because I've been working on a similar problem: Prove that \bar{\ell}₀ = \ell₂ in \ell₂.

With a similar approach of the attempt above I got \bar{\ell}₀ \supseteq \ell₂.

Since \ell₀ \subseteq \ell₂ (for obvious reasons),
\bar{\ell}₀ \subseteq\ell₂ iff \ell₂ = \bar{\ell}₂ (correct me if I'm wrong here).

In that case, \bar{\ell}₀ = \ell₂ in \ell₂.


(I guess it might is trivial that \ell₂ = \bar{\ell}₂ in \ell₂, but Is there an easy way to prove this?)

Any comments?

 

[micromass]

Hmmm.. When is it true that A \subseteq B \Rightarrow \bar{A} \subseteq \bar{B} ?

This is always true. Try to prove it.

(I guess it might is trivial that \ell₂ = \bar{\ell}₂ in \ell₂, but Is there an easy way to prove this?)

If X is a metric space, then \overline{X}=X (closure in X). Try to prove this.

 

[Paalfaal]

Try to prove it.

I certiantly will!


Thank you for your help! Much appreciated